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Umnica [9.8K]
3 years ago
5

How much heat energy is required to melt 75g of ice at 0°C?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
4 0

The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:

Q = N x ΔH

where:

Q  = heat

N  = moles  

ΔH  = enthalpy

In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:


1. Calculate moles of water

2. multiply by the enthalpy of fusion

3. Convert kJ to J.


Step 1 : Calculate moles of water

[ 75g ] x (\frac{1 mol}{18.02g} ) =

Step 2 : Multiply by enthalpy of fusion

Q = N × ΔH  = <em> [ Step 1 Answer ]</em> ×  6.02 =

Step 3 : Convert kJ to J

[ Step 2 Answer ] x (\frac{1000j}{1kJ} ) =

Finally rounding to 2 sig figs (since 34°C has two sig figs) we get


Q Would Equal ____

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Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

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\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

T_2=232.9 \textdegree C

The temperature changes to <u>232.9 degrees Celsius.</u>

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